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03-08-2014, 06:09 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,363

Quote:
 Originally Posted by ScorpioAB From an online study material, an example question under Age-At-Death RV. An age-at-death random variable is modeled by an exponential random variable with PDF f(x) = 0.34e^0.34x; x >= 0. Use the given PDF to estimate Pr(1 < X < 1:02). The answer is just one line: Pr(1 < X < 1.02) 0.02*f(1) = 0.005 Could anyone tell me how this is derived??? Thanks!
This is just probability and calculus, nothing unique to MLC. For any continuous random variable X, the probability that X is between a and b is $\int_a^b f(x)dx$

From calculus, an integral is an area under the graph. As long as a and b are close together and the value of f(x) does not vary much over the integral, that area is pretty close to the area of the trapezoid with corners (a,0), (b,0),(a,f(a)), (b,f(b)) whose area is (b-a)*(f(a)+f(b))/2. That's only an estimate of the area, but for small b-a it's usually a pretty good estimate.

It's not the estimate used here, though. Here's it's even more of an estimate.

Since a and b are close, f(a) and f(b) are close, so f(b) is pretty close to f(a). Therefore (f(a)+f(b)) is approximately equal to f(a)+f(a)=2f(a), and our first estimate is pretty close to (b-a)*2f(a)/2=(b-a)*f(a). That's the estimate the author uses. Thinking geometrically, that's saying the integral is approximately the area of a rectangle with width (b-a) and height f(a).

Other similar estimates would be (b-a)*f(b) or (b-a)*f((a+b)/2). Those are still estimating the integral as the area of a rectangle, just different estimates of the height of the rectangle. If you tried them with your pdf (which is probably missing a - sign), all four estimates would be pretty close.

Quote:
 BTW it's interesting that exam MLC has been revised for several times, from pencil & paper MC to almost CBT, and now it contains both MC and WA.
What is "almost CBT" and when was it that?
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